As a result of this question, I’ve been thinking about the following condition on a topological space Y:

For every topological space X, E⊆X, and continuous maps f,g:X→Y, if E is dense in X, and f and g agree on E (that is, f(e)=g(e) for all e∈E), then f=g.

If Y is Hausdorff, then Y satisfies this condition. The question is whether the converse holds: if Y satisfies the above condition, will it necessarily be Hausdorff?

If Y is not at least T1, then Y does not have the property: if u,v∈Y are such that u≠v and every open neighborhood of u contains v, then let X be the Sierpinski space, X={a,b}, a≠b, with topology τ={∅,{b},X}, E={b}, let f,g:X→Y be given by f(a)=f(b)=v, and g(a)=u, g(b)=v. Then both f and g are continuous, agree on the dense subset E, but are distinct.

My attempt at a proof of the converse assumes the Axiom of Choice and proceeded as follows: assume Y is T1 but not T2; let u and v be witnesses to the fact that Y is not T2, let U_s and V_t be the collection of all open nbds of s that do not contain t, and all open nbds of t that do not contain s, respectively. Construct a net with index set U_s×V_t (ordered by (U,V)≤(U′,V′) if and only if U′⊆U and V′⊆V) by letting y(U,V) be a point in U∩V (this is where AC comes in). Let E={y(U,V)∣(U,V)∈U_s×V_t}, and let X=E∪{s}. Give X the induced topology; let f:X→Y be the inclusion map, and let g:X→Y be the map that maps E to itself identically, but maps s to t.

The only problem is I cannot quite prove that g is continuous; the difficulty arises if I take an open set O∈Vt; the inverse image under g is equal to ((O∩X)−{t})∪{s}, and I have not been able to show that this is open in X.

So:

Does the condition above characterize Hausdorff spaces?

If not, I would appreciate a counterexample. If it

doescharacterize Hausdorff, then ideally I would like a way to finish off my proof, but if the proof is unsalvageable (or nobody else can figure out how to finish it off either) then any proof will do.

Added:A little digging turned up this question raised in the Problem Section of theAmerican Mathematical Monthlyback in 1964 by Alan Weinstein. The solution by Sim Lasher gives a one paragraph proof that does not require one to consider T1 and non-T1 spaces separately.

**Answer**

[This answer moved from this question on Arturo’s advice]

I think the following goes a long way towards proving a converse:

Let (Y,T) be any T1 topological space with at least two points and let a and b be distinct points in Y.

Let X=Y∖{b}.

Let f:X→Y be the inclusion of X in Y.

Let g:X→Y agree with f on X∖{a} and g(a)=b.

Finally, define the topology on X to be the coarsest topology that makes both f and g continuous.

With these assumptions it turns out that X∖{a} is dense in X if and only if a and b do not have disjoint neighbourhoods in Y.

To see that this is true, let us construct a base of the topology on X.

To make f continuous we only need to take the subspace topology. Since

X is open in Y this is S1={G∈T∣b∉G}. To also make g continuous we need to add the open neighbourhoods of b, with b replaced by a. This gives S2={(H∖{b}∪{a}∣H∈T,b∈H}.

Now S1∪S2 is a subbase of the topology on X.

Since S1 and S2 are already closed under finite intersection, and each covers X, we can say that

B={G∩H∣G∈S1,H∈S2} is a base of the topology.

Then (remembering that finite sets are closed in Y) we find that the following are all equivalent:

- X∖{a} is not dense in X
- {a} is open in X
- {a}∈B
- there are G∈S1,H∈S2 such that G∩H={a}
- there are G∈S1,H∈S2 such that a∈G and G∩(H∖{a}∪{b})=⊘
- a and b have disjoint neighbourhoods in (Y,T).

**Attribution***Source : Link , Question Author : Arturo Magidin , Answer Author : Community*